1001 A+B Format(20分)
Calculate a+b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits).
Input Specification:
Each input file contains one test case. Each case contains a pair of integers a and b where $−10^6≤a,b≤10^6$. The numbers are separated by a space.
Output Specification:
For each test case, you should output the sum of a and b in one line. The sum must be written in the standard format.
Sample Input:
-1000000 9
Sample Output:
-999,991
解题思路：
首先看到问题里面a和b的值都只在$[-10^6,10^6]$，因此用整型就足够完成加法运算，因此这里可以直接保存结果为字符串result。接下来只需要倒序遍历字符串，每3个字符就插入1个逗号，需要注意的是，最后一组不能用逗号将符号和数字分离了，比如题目里的输出：$-999,991$，就不能写成$-,999,991$。因此需要加一个判断最开头的一位是否为符号即可。
Code:
1 #include<iostream> 2 #include<string> 3 using namespace std; 4 5 int main(){ 6 int a,b; 7 cin>>a>>skipws>>b; 8 string result = to_string(a+b); 9 int number = 1; 10 int length = result.length(); 11 string comma = ","; 12 for(int i = length-1;i>=1;i--){ 13 if(number % 3 == 0 && isdigit(result[i-1])){ 14 result = result.insert(i,comma); 15 number = 1; 16 } 17 else{ 18 number++; 19 } 20 } 21 cout<<result<<endl; 22 return 0; 23 }
1002 A+B for Polynomials (25 分)
This time, you are supposed to find $A+B$ where $A and $B$ are two polynomials.
Input Specification:
Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial:
$K N_1 a_{N_1} N_2 a_{N_2} … N_k a_{N_k}$
where $K$ is the number of nonzero terms in the polynomial, $N_i$ and $a_{N_i}(i = 1,2,…,K)$ are the exponents and coefficients, respectively. It is given that $1 leq K leq 10, 0 leq N_k < … < N_2 < N_1 leq 1000.$
Output Specification:
For each test case you should output the sum of $A$ and $B$ in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate to 1 decimal place.
Sample Input:
2 1 2.4 0 3.2 2 2 1.5 1 0.5
Sample Output:
3 2 1.5 1 2.9 0 3.2
解题思路：
这道题关键是要读懂题目里面说的是什么，是两个多项式相加，那就是次数不变，系数相加，根据给出的$K$的范围容易得出一开始要设定的数组长度。用数组索引作次数，数组存放的内容做系数，然后相加完以后再倒序遍历即可。
坑点：
测试点2和测试点5：这道题如果在(指数 系数)后面加空格的话，由于你不知道这个多项式的和的结果是否有常数项，如果没有常数项，那就会在Output中多了一个末尾的空格。因此要设置一个是否已经有输出的标识符，判断在输出(指数 系数)时前面是否需要加空格。
测试点6：$A+B=0$，对0要单独作额外的分类即可。
Code:
1 #include<iostream> 2 #include<string> 3 #include<iomanip> 4 using namespace std; 5 double A[1001] = { 0 }; 6 double B[1001] = { 0 }; 7 int main() { 8 int items; 9 cin >> items >>skipws; 10 int times; 11 double coff; 12 for (int i = 0; i < items; i++) { 13 cin >> times >> skipws >> coff >> skipws; 14 A[times] = coff; 15 } 16 cin >> items>>skipws; 17 for (int i = 0; i < items; i++) { 18 cin >> times >> skipws >> coff >> skipws; 19 B[times] = coff; 20 } 21 int number = 0; 22 for (int i = 0; i < 1001; i++) { 23 A[i] += B[i]; 24 if (A[i] != 0) { 25 number++; 26 } 27 } 28 if (number != 0) { 29 cout << number << " "; 30 bool has_been_output = false; 31 for (int i = 1001; i >= 1; i--) { 32 if (A[i] != 0) { 33 if (!has_been_output) { 34 cout << i << " " << fixed << setprecision(1) << A[i]; 35 has_been_output = true; 36 } 37 else { 38 cout << " " << i << " " << fixed << setprecision(1) << A[i]; 39 } 40 } 41 } 42 if (A[0] != 0) { 43 cout << " " << 0 << " " << fixed << setprecision(1) << A[0] << endl; 44 } 45 } 46 else { 47 cout << number << endl; 48 } 49 return 0; 50 }
(未完待续)